Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		double[] p=new double[n];
		for (int i=0;i<n;++i)
			p[i]=in.nextDouble();
		double[][] dp=new double[n+1][n+1];
		dp[0][0]=1;
		for (int i=0;i<=n;++i)
			for (int j=0;j<=n;++j) if (i>0||j>0) {
				if (i+j>n) break;
				int t=i+j-1;
				if (i>0) dp[i][j]+=dp[i-1][j]*p[t];
				if (j>0) dp[i][j]+=dp[i][j-1]*(1-p[t]);
			}
		double res=0;
		for (int i=n/2+1;i<=n;++i)
			res+=dp[i][n-i];
		System.out.printf("%.12f\n",res);
	}
}